and the generalization of $\zeta (2k)$, my perception of $\pi$ has changed. I used to think of it as rather obscure and purely geometric (applying to circles and such), but it seems that is not the case since it pops up in things like this which have no known geometric connection as far as I know. What are some other cases of $\pi$ popping up in unexpected places, and is there an underlying geometric explanation for its appearance?

In other words, what are some examples of $\pi$ popping up in places we wouldn't expect?

$\begingroup$@user92774 The explanation actually would make a nice addition to my answer below. It's a special case of the central limit theorem. Binomial distributions tend toward the normal distribution for large $n$, which relates the $n!$ in the normalization factor of the binomial distribution to the $\sqrt{2\pi}$ in the normal distribution's normalization factor.$\endgroup$
– David HFeb 25 '14 at 3:07

2

$\begingroup$@user92774 The $\sqrt{\pi}$ in the Stirling's formula comes from the Laplace's approximation of a uni-modal function by the Gaussian function which integrates to $\sqrt{\pi}$. So the real question is why it integrates to that.$\endgroup$
– VadimFeb 25 '14 at 3:08

$\begingroup$I wonder why this interesting question "where does $\pi$ pop up unexpectedly" has been misinterpreted as "what are fancy formulas for $\pi$" (except for two answers).$\endgroup$
– Martin BrandenburgMar 1 '14 at 12:43

27 Answers
27

What are some interesting cases of $\pi$ appearing in situations that are not geometric ?

None! :-) You did well to add “do not seem” in the title! ;-)

All $\zeta(2k)$ are bounded sums of squares, are they not ? And the equation of the circle, $x^2+y^2=$$=r^2$, also represents a bounded sum of squares, does it not ? :-) Likewise, if you were to read a proof of why $\displaystyle\int_{-\infty}^\infty e^{-x^2}dx=\sqrt\pi$ , you would see that it also employs the equation of the circle! $\big($Notice the square of x in the exponent ?$\big)$ :-) Similarly for $\displaystyle\int_{-1}^1\sqrt{1-x^2}=\int_0^\infty\frac{dx}{1+x^2}=\frac\pi2$ , both of which can quite easily be traced back to the Pythagorean theorem. The same goes for the Wallis product, whose mathematical connection to the Basel problem is well known, the former being a corollary of the more general infinite product for the sine function, established by the great Leonhard Euler. $\big($Generally, all products of the form $\prod(1\pm a_k)$ are linked to sums of the form $\sum a_k\big)$. It is also no mystery that the discrete difference of odd powers of consecutive numbers, as well as its equivalent, the derivative of an odd power, is basically an even power, i.e., a square, so it should come as no surprise if the sign alternating sums$(+/-)$ of the Dirichlet beta functionalso happen to depend on $\pi$ for odd values of the argument. :-) Euler's formula and his identity are no exception either, since the link between the two constants, e and $\pi$, is also well established, inasmuch as the former is the basis of the natural logarithm, whose derivative describes the hyperbola$y=\dfrac1x$, which can easily be rewritten as $x^2-y^2=r^2$, following a rotation of the graphic of $45^\circ$. As for Viete's formula, its geometrical and trigonometrical origins are directly related to the half angle formula known since before the time of Archimedes. Etc. $\big($And the list could go on, and on, and on $\!\ldots\!\big)$ Where men see magic, math sees design. ;-) Hope all this helps shed some light on the subject.

Suppose we iterate the function $f(z)=z^2+c$ starting at $z_0=0$. For example, if $c=1/4$, the first few terms are
\begin{align}
z_0&=0 \\
z_1&=0^2+1/4=1/4\\
z_2&=(1/4)^2+1/4=5/16
\end{align}
It can be shown that the sequence converges slowly up to $1/2$. On the other hand, if $c=1/4+\delta$, where $\delta>0$ (no matter how small), then the sequence diverges to $\infty$. This corresponds to the fact that $c=1/4$ is on the boundary of the Mandelbrot set.

We now ask the following: given $\delta>0$, how many iterates $N$ does it take until $z_N>2$? Here are the answers for several choices of $\delta$:

$\begingroup$@Lucian Thank you for the generalization. There is something that I don't understand : are you saying that the result I gave can been prove geometrically?$\endgroup$
– user37238Mar 4 '14 at 12:25

1

$\begingroup$It ultimately lies on a geometric and trigonometric foundation. (The only result in this entire thread that -to my knowledge- does not possess a geometric or trigonometric interpretation, is Ramanujan's $\pi$ formula, all others being ultimately linked or connected with these two closely-related fields).$\endgroup$
– LucianMar 4 '14 at 13:01

$\begingroup$At some deep level this is how you get $\sqrt{\pi}$ in Stirling's formula.$\endgroup$
– VadimFeb 25 '14 at 3:11

7

$\begingroup$However, the Gaussian is quite geometric, and the Gaussian integral is not a result of statistical analysis of data, but rather statistical analysis of data piggybacks on the geometric properties of the Gaussian.$\endgroup$
– EmilyFeb 25 '14 at 21:46

$\begingroup$@Emily "However, the Gaussian is quite geometric": can you substantiate this ?$\endgroup$
– Yves DaoustFeb 2 '17 at 15:40

$\begingroup$Consider computing its square (for which the answer is $\pi$), and the circle becomes clear by looking at the exponent. If it's not obvious, convert to polar coordinates.$\endgroup$
– Glen_bApr 2 at 7:44

$\begingroup$I like that one too! I believe, though, that it can be used to find hexadecimal digits of pi without calculating the all previous. The relationship between base 16 and base 2 allows you to do the same with binary but I don't believe it works in base 10.$\endgroup$
– Mark McClureFeb 25 '14 at 12:02

$\begingroup$@MarkMcClure Yes, they are hex digits. A bonus is that it's only practical to use these formulas with computers which can work with binary very efficiently.$\endgroup$
– qwrFeb 26 '14 at 2:09

$\begingroup$This is called a spigot algorithm. IIRC there was a version for base 10 but I can't find it now.$\endgroup$
– DanielVFeb 26 '14 at 6:15

$\begingroup$That it is very close to an integer. In fact we have $262537412640768743.9999999999992 = (640319.9999999999999999999999993\ldots)^3 + 744$ but this is of course not easy to see from the number alone, whereas the other fact is. Likewise, we have $e^{\pi\sqrt{67}} = (5279.99999999999998\ldots)^3 + 744$, $e^{\pi\sqrt{43}} = (959.99999999991\ldots)^3 + 744$ and a few more like that, in descending order of spectacularity.$\endgroup$
– doetoeOct 15 '14 at 7:37

1

$\begingroup$+1. Is there any reason or pattern to these numbers?$\endgroup$
– JoaoOct 16 '14 at 3:02

3

$\begingroup$@Joao Yes there is. It has to do with the fact that $\Bbb Q(\sqrt{-163})$, $\Bbb Q(\sqrt{-67})$, $\Bbb Q(\sqrt{-43})$ have unique factorization. The values $e^{\sqrt{163}\pi}$ etc. (essentially) are the most significant term of the Fourier coefficients of the j-invariant as a complex function on the complex upper half plane evaluated in $\tau = \frac12 + \frac12\sqrt{-163}$. Writing $q = e^{2\pi i\tau}$ the Fourier series of $j$ starts like $q^{-1} + 744 + 196884q + \cdots$, and for theoretical reasons this has to be an integer for these values.$\endgroup$
– doetoeOct 16 '14 at 7:35

$\begingroup$$q$ is very small for the $\tau$ I mentioned, namely $q = -e^{-\sqrt{163}\pi}$, so that $q^{-1}$ should be close to an integer. Note that much more can be said.$\endgroup$
– doetoeOct 16 '14 at 7:40

I think it is the simplest form that may have some geometric interpretation. Actually, once I discovered this expression in high school, I spent some time thinking about what the geometric explanation for this might be, but don't think came up with something nice.

$\begingroup$Oh yeah, that's a fun one. When I'm really bored in Calc class (I already know all of the material but have to take the class anyway...) I would compute $\pi$ using this formula. The incredibly slow convergence of this formula made my unproductive Calc days even more unproductive! (I almost managed to calculate it to three decimal points of accuracy one day...)$\endgroup$
– MCTFeb 25 '14 at 3:08

1

$\begingroup$Yes, it is slow. But if you are looking for something that might have a geometric explanation, this is the closest I can think of.$\endgroup$
– VadimFeb 25 '14 at 3:10

3

$\begingroup$isn't this the just the series expansion of $\arctan 1$?$\endgroup$
– ratchet freakFeb 26 '14 at 13:17

$\begingroup$@ratchetfreak Yes, but I do not know a geometric explanation for the series expansion of $\arctan$.$\endgroup$
– VadimFeb 26 '14 at 17:42

The "Buffon's needle experiment" says that if a needle of length $l$ is tossed on a paper ruled with lines with $d$ distance apart and equidistant from each other and also $l<d$, then the probability of the needle crossing one of the ruled line is

$${P=\large \frac{2l}{\pi d}}$$

Consequently, if $l=d$, then $\pi$ can be calculated as

$\pi=\Large \frac{2}{P}$, where $P=\Large \frac{\text{number of tosses when the needle crosses on of the lines}}{\text{Total number of tosses}}$

In 1901 the Italian mathematician Mario Lazzarini tried this with 3,408 tosses of the needle and got $\pi = 3.1415929$.

$\begingroup$(Another unexpected appearance of $\pi$, but much more well-known that the above one, is : the probability of a random integer to be square-free is $6 / \pi^2$)$\endgroup$
– WatsonMay 28 '18 at 21:11

$\begingroup$Another example: what is the average distance between two random points in the unit cube? If you think that this is unrelated to $\pi$ and to logarithms, see here.$\endgroup$
– WatsonNov 29 '18 at 17:34

$\begingroup$Not fair - that has a rather obvious geometric interpretation.$\endgroup$
– nbubisFeb 25 '14 at 3:23

$\begingroup$I have no idea what that would be.$\endgroup$
– Martin ArgeramiFeb 25 '14 at 3:30

2

$\begingroup$I drew your example on a large poster and hung it on a wall during a math club event. :)$\endgroup$
– neofoxmulderFeb 25 '14 at 6:39

1

$\begingroup$The really interesting break from geometry here occurs at the point where you prove that the anti-derivative of $\arctan$ can be expressed without any trig functions. This is just a corollary.$\endgroup$
– Jack MFeb 25 '14 at 12:03

4

$\begingroup$The equivalence of areas can be shown without trig and calculus, by mere triangles similarity in the greek style. If this helps: i.imgur.com/amN1Exb.png$\endgroup$
– leonbloyMar 20 '14 at 15:04

$\begingroup$for the first $\frac{1}{{2n\choose n}}$ the Central Binomial Coefficients an for the second $\frac{n+1}{{2n\choose n}}$ the Catalan Numbers, reciprocals both$\endgroup$
– janmarqzFeb 25 '14 at 4:16

$\begingroup$In general, $~\displaystyle\sum_{n=1}^\infty\dfrac{(2x)^{2n}}{\displaystyle{2n\choose n}~n^2}~=~2\arcsin^2x.~$ By repeatedly differentiating with regard to $x,~$ we can deduce the general formulas behind the two provided identities.$\endgroup$
– LucianFeb 2 '17 at 18:14

$\begingroup$This still kind of follows from $e^{\frac{1}{2} i \pi} = i$ which has an obvious geometric interpretation.$\endgroup$
– ThomasFeb 25 '14 at 14:29

1

$\begingroup$An imaginary number raised to an imaginary power can be a real number and $ \pi $ shows up. I would think that is a very strange place to find $ \pi $ :)$\endgroup$
– neofoxmulderMar 1 '14 at 12:54

$\begingroup$The mentioning of the fundamental group is a joke, right?$\endgroup$
– Martin BrandenburgMar 1 '14 at 12:41

$\begingroup$@MartinBrandenburg: user92774 never said he was talking about the constant defined as the ratio of a circle's circumference to its diameter :-) I thought about adding more from this list, but that would be too much. Sometimes I have the feeling that people mix up the representant ($\pi$) with what they represent (the ratio of a circle's circumference to its diameter). That's why I've added the fundamental group.$\endgroup$
– Martin ThomaMar 1 '14 at 13:21

2

$\begingroup$I wouldn't say that the fundamental group "do not seem geometric", as $\pi_1$ measures the loops/circles. For the Gauss-Bonet, +1 :)$\endgroup$
– Peter FranekJun 27 '14 at 11:35

$\begingroup$@MartinThoma Hey, this is OP (I've since changed my name). Just learned about the fundamental group and couldn't help but think about this post! :) That said I don't think people "mix up the representant with what they represent" -- isn't the point of notation so that I can write $\pi$ and not "the ratio of the circumference of a circle and the diameter of a circle of same radius", which is pretty reminiscent of ancient math textbooks...$\endgroup$
– MCTNov 1 '16 at 0:10

I'm not sure what you mean by "geometric". If you mean ratios of circumference to diameter and such, then I think this might fit your criterion. :) Nevertheless, this is such a beautiful formula that I felt it was worth mentioning.

$\begingroup$Well, to me there is a clear connection with complex numbers and geometry: $e^{ix}$ is the graph of a circle of radius $1$ on the complex plane.$\endgroup$
– MCTFeb 25 '14 at 2:42

2

$\begingroup$Sure, the image of the set $[0,2\pi),$ under the map $e^{ix},$ is the unit circle. The mentioned identity says that the $\pi \mapsto -1.$ But it's still not so much the "circumference to diameter" business.$\endgroup$
– RaghavFeb 25 '14 at 2:47

$\begingroup$Well, it sort of is; the statement is that, if you want to trace out a circle of radius $r = 1$ (i.e., $d = 2$), then you need to consider $x$-values from $0$ to $2\pi$ to do it (i.e., $C = 2\pi$). The mere numerical coincidence doesn't prove it—one needs also to know that $x \mapsto e^{i x}$ traces out the unit circle at unit speed—but the connexion is surely there.$\endgroup$
– LSpiceAug 2 '15 at 21:43

This is too long for a comment, but this regards the $\frac{\pi}{4}$ series mentioned earlier.

Consider a unit square and a quarter of the unit circle. Cut the square down the diagonal; half of the arc of that circle will equal $\frac{\pi}{4}$. Split $AB$ into $n$ equal pieces. Then it can be shown that $$\lim_{n \to \infty} \displaystyle \sum_{r = 1}^{n} \frac{\frac1n}{1 + (\frac rn)^2} = \frac{\pi}{4}$$

In statistics, if hypothesis test A requires sample size $n_A$ to attain a certain power $\beta$, and hypothesis test B requires sample size $n_B$, we say that the relative efficiency of test A with respect to test B is $\frac{n_B}{n_A}$. Effectively this tells us how efficiently the tests make use of the information in the samples to draw inferences about the population. If Test A requires only half the sample size that Test B did, then we say it has twice the efficiency. The asymptotic relative efficiency considers how the relative efficiency behaves for increasingly large sample sizes.

Two frequently-used hypothesis tests are the Student's t-test for independent samples (which assumes data are drawn from a normal distribution), and the Wilcoxon-Mann-Whitney U test (which is a non-parametric test: it does not assume a normal distribution). If the data are actually drawn from an exponential distribution, the U test "wins": compared to the t-test it has an ARE of 3, i.e. it is 3 times as efficient.

But if the data are drawn from a normally distributed population, then Student's t-test is on home turf and its power benefits from the data fulfilling its assumption of normality. The U test is now at a slight disadvantage. Compared to the t-test, its ARE drops to $\frac{3}{\pi} \approx 0.955$.

$\begingroup$This is David H's answer in disguise: the ARE is algebraically related to (a) $e^0=1$ and (b) $\int_\mathbb{R}e^{-x^2}dx=\Gamma(1/2)=\sqrt{\pi}$. The appearance of $\pi$ is due to the duplication formula for the Gamma function, $\Gamma(z)\Gamma(1-z)=\pi\csc(\pi z)$, exhibiting $\Gamma$ as a kind of "square root" of a trig function. This is because $\Gamma(z)=\Gamma(z+1)/z$ implies $\Gamma$ has (simple) poles at $0,-1,-2,\ldots$, whence $\Gamma(z)\Gamma(1-z)$ has poles at $\mathbb Z$, strongly suggesting periodicity--which is why $\pi$ should show up!$\endgroup$
– whuberDec 29 '14 at 16:56

Stirling's Formula : $n!\sim (n/e)^n\sqrt {2\pi n}.$ It is fairly easy to show that $n!\sim (n/e)^n\sqrt {kn}$ for some $k$ but to show $k=2\pi$ is more subtle, and is basically the same as finding the Wallis Product for $\pi.$ Which Wallis did before Newton, that is, before calculus.

For each $n\in\mathbb N$, let $P_h(n)$ be the number of primitive Pythagorian triples whose hypothenuse is smaller than $n$ and let $P_p(n)$ be the number of primitive Pythagorian triples whose perimeter is smaller than $n$. In 1900, Lehmer proved that

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